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How can I prove that the number formed by concatenating the primes in order i.e. $0.235711131719...$ is irrational.

I know I have to demonstrate that it has no period, but I'll be so thankful if someone can explain very clear, including all cases.

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Why is proving that it has no period sufficient? – Christopher A. Wong May 2 '14 at 18:33
This is the Copeland-Erdos constant. – Lucian May 2 '14 at 19:10
up vote 9 down vote accepted

We use Dirichlet's Theorem on primes in arithmetic progressions. Let $A_k$ be the number whose decimal expansion consists of $k$ consecutive $1$'s. The numbers $A_k$ and $10^{k+1}$ are relatively prime. It follows by Dirichlet's Theorem that there are infinitely many primes of the shape $A_k +n\cdot 10^{k+1}$. Such a prime has $k$ consecutive $1$'s at the left end of its decimal expansion.

Thus your number has strings of consecutive $1$'s of arbitrary length. By another simpler application of Dirichlet's Theorem, the number has non-$1$'s arbitrarily far in its decimal expansion. Thus the decimal expansion cannot be ultimately periodic.

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$0.23571113...$ is known as the Copeland–Erdős constant and it is given by: $$ \displaystyle \sum_{n=1}^\infty p_n 10^{-\left( n + \sum\limits_{k=1}^n \left\lfloor \log_{10}{p_k}\right\rfloor \right)} .$$ Here are two proofs given in Hardy's book An Introduction to the Theory of Numbers (p. 113) that shows that it is irrational:

$\phantom{}1)$ Let us assume that any arithmetical progression of the form $$k10^{\,s+1}+1\quad(k=1,2,3,\ldots)$$ contains primes. Then there are primes whose expressions in the decimal system contain an arbitrary number $s$ of $\rm O$ 's, followed by a $1$. Since the decimal contains such sequences, it does not terminate or recur.

$\phantom{}2)$ Let us assume that there is a prime between $N$ and $10N$ for every $N \geqslant 1$. Then, given $s$, there are primes with just $s$ digits. If the decimal recurs, it is of the form $$... a_ 1 a_ 2 ...a_ k |a_ 1 a_ 2 ...a_ k |...$$ the bars indicating the period, and the first being placed where the first period begins. We can choose $l\gt1$ so that all primes with $s = kl$ digits stand later in the decimal than the first bar. If $p$ is the first such prime, then it must be of one of the forms $$p = a_ 1 a_ 2 ...a_ k |a_ l a_ 2 ...a _k| ...|a _1 a_ 2 ...a_ k$$ or $$p = a_ {m+1} ..a_ k| a _1 a_ 2 ...a_ k| ...|a_ l a _2 ...a _k| a _l a _2 ...a_ m$$ and is divisible by $a$, $a_2 ...a_k$ or by $a_{m+1} ..a_k a$, $a_2 ...a_m$ : a contradiction.

The first proof follows from a special case of Dirichlet's theorem and the second from a theorem stating that for every $N\geqslant1$ there is at least one prime satisfying $N < p\leqslant2N$. It follows, a fortiori, that $N < p < 10N$.

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Suppose this were to start repeating after some point with a period $N$. Beyond this point, for any $k$ there are at most $N$ different strings of length $k$ appearing in the repeating part. So it is enough to show that for large values of $k$ we have more than $N$ primes with $k$ digits. The prime number theorem tells us to expect the number of primes with $k$ digits to be on the order of $\frac{10^k}{k}$, so in particular for $k >> 0$ we can make this bigger than any $N$.

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Not really formal but should point you in the right direction. Well let's suppose there is a period. Then the period would be $n$ digits long. Now I can find a prime that is longer than $n$ and therefore I cannot have a finite period.

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This really isn't quite enough, here's a silly example of why not: Suppose we had a period 1111... we would still need to rule out having only primes of the form 1111...1 from some point on. – Nate May 2 '14 at 19:11
This does not address the question, really. And the reasoning is faulty. If the period is $13$, you could still have a prime such as $131$. – Andrés Caicedo May 2 '14 at 19:15

Suppose this number is eventually periodic with a period length of $L$. Then eventually we have infinitely many primes in the sequence such that $P > 10^L$, and for some $P$, $10^{nL} > P \geq 10^{nL - 1}$ (in other words we can find a prime whose digit length is nL, a multiple of L)$^{[1]}$. Then since P has a digit length which is a multiple of L, we can write:

$$P = R \cdot 10^{L(n - 1)} + R \cdot 10^{L(n - 2)} + \cdots + R$$

Where $R$ is the repeating sequence. But then $R|P$, and $P$ is not prime, a contradiction.

[1] This is justified by Bertrand's postulate, see the comments in this answer.

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No. The final bit does not have to be $R$, it could be a proper initial segment of $R$. For instance, if $R=13$, $P$ could be $131$. – Andrés Caicedo May 2 '14 at 19:03
The length of P is chosen to be a multiple of the length of R, so your counter-example doesn't work. It's true that $R$ doesn't have to be the original repeating sequence, but once the sequence is repeated, you can choose any point as the starting point, and I'm choosing $R$ such that $R$ is the repeating sequence that starts exactly when $P$ starts. Since $P$ has length $nL$ this isn't a problem. – FlagCapper May 2 '14 at 19:08
How do you know that there are primes of that length? – Andrés Caicedo May 2 '14 at 19:10
Bertrand's postulate for one thing. It says that there is always a prime between $n$ and $2n$, and so certainly always a prime between $n$ and $10n$, which is what I'm using in this proof. The prime number theorem or other results on the distribution of primes are also enough. – FlagCapper May 2 '14 at 19:12
You may want to edit your answer, so these issues are explicitly addressed there and not hidden in the comments. (Bertrand's postulate is much more elementary than the prime number theorem, so I suggest using it in the argument.) – Andrés Caicedo May 2 '14 at 19:17

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